C did the most effort, but A walked the longest, so a reasonable answer is that they all did the same amount of work, as B simply mediates.
Answer:


Explanation:
= Force on front wheels
= Force on rear wheels
Distance between CG and rear wheel = 3.99-0.622 = 3.368 m

As the forces are conserved we have

On rear wheels

The force on each rear whees is 
Force on each front wheel is 
The electric flux on a surface of area A in a uniform electric field E is given by
φ = EA cos θ
where θ = the angle between the directions of E and a normal vector to the surface of the area.
See the diagram shown below.
When the electric field is perpendicular to the surface, then θ = 0°, and
φ = EA cos(0°) = (6.20 x 10⁵ N/C)*(3.2 m²) = 1.984 x 10⁶ (N-m²)/C
When the electric field is parallel to the area, then θ = 90°, and
φ = EA cos(90°) = 0
Answer:
(a) 1.984 x 10⁶ (N-m²)/C
(b) 0
Answer:
Explanation:
For a string of length L clamped down on both ends , for fundamental mode of vibration
2 . λ / 4 = L
= λ / 2 = L
λ = 2 L
For other modes , the formula is
n . λ/2 = L where n is an integer .
λ = 2L / n
if n = 6
λ = 2L / 6
= L/3
So wavelength = L/3 is possible .
Answer:
μ = 0.179
Explanation:
For this exercise we use the rotational equilibrium condition, where we set a reference system at the top of the ladder where it is in contact with the parent, we will assume the counterclockwise rotations as positive
∑ τ = 0
W_stairs x/2 + W_man x’+ fr y - N x = 0
fr y = -W_stairs x/2 - W_man x’ + N x
let's use trigonometry for distances
* the man up the ladder 78%
l ’= 0.78 L
cos 59 = x ’/ l’
x’= 0.78 L cos 59
* the horizontal distance
cos 59 = x / L
x = L cos 59
* vertical distance
sin 59 = Y / L
y = L sin 59
we substitute
fr L sin 59 = -W_stairs Lcos59 / 2 - W_man 0.78 L cos59 + N L cos59
fr sin 59 = - ½ W_stairs cos 59 - 0.78 W_man cos 59 + N cos 59
fr = ctan 59 (N - ½ W_stairs - 0.78 W_man) (1)
Let's write the translational equilibrium equations
Y axis
N -W_stairs - W_man = 0
N = W_stairs + W_man
N = 85 + 21.5 9.8
N = 295.7 N
we substitute in 1
fr = ctg 59 (295.7 - ½ 85 - 0.78 21.5 9.8)
fr = 0.6 (295.7 - 42.5 - 164.346)
fr = 52.89 N
the expression for the friction force is
fr = μ N
μ = fr / N
μ = 52.89 / 295.7
μ = 0.179